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Because the function is monotonic this locates distinct rational number in each discontinuity. The open intervals (supL,infU), at the points of discontinuity, are disjoint because the function is monotonic. A set of rationals is countable,so the set of discontinuities of a monotonic function is countable.

Alright so "this locates distinct rational number in each discontinuity" is kinda wierd so why not irrational and irrational are uncountable so this tells discontinuities of the function is uncountable.

  • Since this is a very well known result whose proof can be found in many places on the internet and in many books, and since I have no idea what you mean by "many proof which has only Rational mentioned", I would be interested in a specific example of a proof that you think has this shortcoming. If a book, give the specific book, edition, page number; if on the internet, give a link. – Dave L. Renfro Feb 22 '20 at 20:58
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    You can create an injective function from the set of discontinuities to the irrational numbers but you cannot prove it is surjective, therefore you cannot conclude the set of discontinuities of the function is uncountable. – Miguel Feb 22 '20 at 21:09
  • I think what you're referring to is a way of proving countability of a set by establishing a 1-1 mapping (an injection) to the set of rational numbers. Thus, one way to show a certain set $E$ is countable is to show there exists a 1-1 function $f:E \rightarrow {\mathbb Q}.$ Usually one uses $\mathbb N$ instead of ${\mathbb Q},$ but $\mathbb Q$ works as well and sometimes is more natural to use in certain situations. (As is evident by all my wording changes, I frequently get confused in how to write the correct direction!) – Dave L. Renfro Feb 22 '20 at 21:10
  • @Miguel Thanks For Making it less painful. –  Feb 22 '20 at 21:12

2 Answers2

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Here is another approach which you may find useful.

Let $f$ be a monotonic function on a closed and bounded interval $[a, b] $. Then the set $D$ of discontinuities of $f$ on $[a, b] $ is countable.

Let's assume $f$ is increasing on $I$. If $f(a) =f(b) $ then $f$ is constant and therefore continuous so that $D$ is empty. Let's assume $f(a) <f(b) $. Since $f$ is increasing it may possess only jump discontinuities and the right hand limit of $f$ will be greater than its left hand limit at each point of its discontinuity. Let the difference of these limits at point $c$ be called jump at $c$. Consider the set $D_n, n\in\mathbb {N} $ defined by $$D_n=\{x\mid x\in[a, b], \text{ jump of } f\text{ at } x> 1/n\}$$ The sum of jumps of $f$ can't exceed $f(b) - f(a) $ and each jump at points of $D_n$ exceeds $1/n$ and hence the number of points in $D_n$ must be less than $n(f(b) - f(a))$. Thus each $D_n$ is finite and since $D=\cup_{n=1}^{\infty}D_n$ it follows that $D$ is countable.

The extension to open interval $(a, b) $ can be done by noting that $$(a, b) =\bigcup_{i=1}^{\infty} [a+1/n,b-1/n]$$ and the similar argument can be used to deal with $[a, b) $ or $(a, b] $.

The extension to unbounded intervals follows from the fact any unbounded interval including the whole set $\mathbb{R} $ can be written as a countable union of bounded intervals like $$\mathbb{R} =\bigcup_{n=1}^{\infty} [-n, n] $$

Paramanand Singh
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Let $f:\mathbb{R} \to \mathbb{R}$ be an increasing function and $D$ be the set of points where $f$ is discontinuous.

Since the domain of $f$ is $\mathbb{R}, f$ can only have jump discontinuities.

Let $x \in D$, we have $f(x^-)<f(x^+)$, therefore there exists a rational number $a_x$ such that $f(x^-)<a_x<f(x^+)$. We thus have an injective function from $D$ to $\mathbb{Q}$, because if $x< y$ then it is easy to see that $a_x< a_y$. Since $\mathbb{Q}$ is countable the result follows.

The proof for decreasing functions is analogous.

Miguel
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  • Why can't we have function from D to irrationals? Isn't function maps Reals to reals? –  Feb 22 '20 at 21:08
  • @2Balls1Bat: we can have a map from $D$ to irrationals in the same way. But that proves that the cardinality of $D$ is less than or equal to that of irrationals. It does not prove that the cardinality is necessarily equal. In case of rationals the same situations holds and we can have $D$ finite and thus of less cardinality than $\mathbb{Q} $. Thus we have both $|D|\leq|\mathbb{Q}| $ and $|D|\leq|\mathbb {R} \setminus \mathbb {Q} |$. From these it follows that $D$ is countable. – Paramanand Singh Feb 23 '20 at 04:12
  • @2Balls1Bat: Perhaps you were thinking that the argument in this answer gives equality of cardinality of the sets involved. But that's not the case. – Paramanand Singh Feb 23 '20 at 04:14